Reversible reactions and equilibrium Revision Notes

A reversible reaction is a chemical reaction where the reactants form products, which can themselves react to reform the original reactants. This means the reaction can proceed in both the forward and reverse directions under suitable conditions.

Unlike irreversible reactions, where products do not convert back to reactants, reversible reactions reach a state where both forward and reverse reactions occur simultaneously.

An example is the reaction between nitrogen and hydrogen gases to form ammonia, where ammonia can decompose back into nitrogen and hydrogen.

• Reactions proceed in both forward and reverse directions.
• Products can convert back into reactants.
• Reversible reactions are dynamic and can reach equilibrium.

The symbol ⇌ is used in chemical equations to indicate that the reaction is reversible. It replaces the single arrow (→) used for irreversible reactions.

For example, the reversible reaction between nitrogen and hydrogen to form ammonia is written as:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (g = gas) This shows that ammonia can decompose back into nitrogen and hydrogen under certain conditions.

• ⇌ indicates a reversible reaction.
• Used in balanced chemical equations.
• Shows that both forward and reverse reactions occur.

In reversible reactions, the forward reaction converts reactants into products, while the reverse reaction converts products back into reactants.

Initially, the forward reaction rate is higher because there are more reactants. As products form, the reverse reaction rate increases.

Eventually, the rates of the forward and reverse reactions become equal, leading to a dynamic equilibrium where concentrations of reactants and products remain constant (though this is beyond the current objective).

• Forward and reverse reactions occur simultaneously.
• Reaction rates change over time.
• Products can revert to reactants.

1. Formation of ammonia (Haber process): N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

2. Hydration of copper(II) sulfate: CuSO₄(s) + 5H₂O(l) ⇌ CuSO₄·5H₂O(s)

3. Dissociation of cobalt(II) chloride: CoCl₂·6H₂O(s) ⇌ CoCl₂(s) + 6H₂O(g) (colour changes indicate reversibility)

• Reactions involving gases and solids can be reversible.
• Physical changes like hydration/dehydration are reversible.
• Colour changes often indicate reversible reactions.

Understanding reversible reactions is essential for studying chemical equilibrium, reaction yields, and industrial processes.

The ⇌ symbol helps identify reactions where conditions can shift the balance between reactants and products.

Recognising reversibility aids in predicting reaction behaviour and optimising conditions in chemical manufacturing.

• Key to understanding equilibrium.
• Important in industrial chemistry.
• Helps predict reaction outcomes.

Certain ionic compounds exist in hydrated and anhydrous forms. The hydrated form contains water molecules incorporated into the crystal lattice, known as water of crystallisation. These compounds can undergo reversible reactions where heating removes water (dehydration), and adding water restores the hydrated form (hydration).

These reversible reactions are influenced by conditions such as temperature and the presence of water. Understanding these changes is important in practical chemistry and industrial processes.

• Hydrated compounds contain water molecules in their crystalline structure.
• Anhydrous compounds lack these water molecules.
• Heating can remove water, causing dehydration.
• Adding water can restore hydration.

Copper(II) sulfate pentahydrate, CuSO₄·5H₂O, is a blue crystalline solid containing five water molecules per formula unit. When heated, it loses water molecules and converts to anhydrous copper(II) sulfate, CuSO₄, which is white or greyish-white.

This process is reversible: heating drives the reaction towards dehydration, removing water, while adding water reverses it.

The reaction can be represented by the balanced equation: CuSO₄·5H₂O (s) ⇌ CuSO₄ (s) + 5H₂O (g)

• Heating CuSO₄·5H₂O causes it to lose water and change from blue to white.
• The water lost is in the form of water vapour.
• The reaction is reversible; adding water to anhydrous CuSO₄ restores the blue hydrated form.
• Heating must be controlled to avoid decomposition beyond dehydration.

Anhydrous cobalt(II) chloride, CoCl₂, is a blue solid. When water is added, it forms the hydrated compound cobalt(II) chloride hexahydrate, CoCl₂·6H₂O, which is pink.

This hydration reaction is reversible: removing water by heating converts the pink hydrated form back to the blue anhydrous form.

The balanced reversible reaction is: CoCl₂ (s) + 6H₂O (l) ⇌ CoCl₂·6H₂O (s)

• Adding water to blue anhydrous CoCl₂ produces pink hydrated CoCl₂·6H₂O.
• Heating pink CoCl₂·6H₂O removes water and returns it to blue anhydrous CoCl₂.
• Colour changes provide visual evidence of hydration and dehydration.
• The reaction involves incorporation or loss of water molecules in the crystal lattice.

The direction of these reversible reactions depends on temperature and water availability. Increasing temperature favours the endothermic dehydration reaction, shifting equilibrium to the right (towards anhydrous form and water vapour).

Adding water shifts the equilibrium to the left, favouring the hydrated form.

These changes illustrate Le Chatelier's principle: the system adjusts to oppose changes in conditions, such as temperature or concentration of water.

• Heating supplies energy to break bonds between water molecules and ionic lattice.
• Removing water vapour drives dehydration forward.
• Adding water increases concentration of reactant, favouring hydration.
• Colour and physical state changes indicate reaction direction.

In the laboratory, heating hydrated copper(II) sulfate causes a colour change from blue to white as water is lost. The white anhydrous powder can be rehydrated by adding water, restoring the blue colour.

Similarly, cobalt(II) chloride changes from pink hydrated crystals to blue anhydrous powder on heating, and reverses on adding water.

These reversible reactions are useful for testing for water presence and demonstrate equilibrium concepts in solid-state chemistry.

• Use a crucible and Bunsen burner to heat hydrated crystals gently.
• Observe colour change and note physical state before and after heating.
• Add distilled water carefully to anhydrous powder to observe rehydration.
• Repeat to confirm reversibility and consistency of observations.
• Ensure safety by avoiding overheating and inhalation of vapours.

A reversible reaction is a chemical reaction where the reactants form products, which can themselves react to reform the original reactants. This means the reaction can proceed in both forward and reverse directions.

A closed system is one where no substances can enter or leave during the reaction. This containment ensures that the total amount of reactants and products remains constant, allowing equilibrium to be established.

For example, the reversible reaction between nitrogen and hydrogen gases to form ammonia is represented as: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). In a closed container, no gases escape or enter, so the system is closed.

• Reversible reactions proceed in both forward and reverse directions.
• Closed systems prevent exchange of matter with the surroundings.
• Equilibrium can only be established in a closed system.

At the start of a reversible reaction, only reactants are present, so the forward reaction rate is at its maximum, while the reverse reaction rate is zero because no products exist yet.

As products form, the reverse reaction begins, increasing its rate. Meanwhile, the forward reaction rate decreases because reactant concentrations fall.

Eventually, the forward and reverse reaction rates become equal. This dynamic balance means both reactions continue to occur but at the same rate, so there is no net change in concentrations.

• Forward reaction rate starts high and decreases over time.
• Reverse reaction rate starts at zero and increases over time.
• Equilibrium is reached when forward and reverse rates are equal.

Chemical equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction.

At equilibrium, the concentrations of reactants and products remain constant over time, although both reactions continue to occur simultaneously.

This state is dynamic, not static, meaning molecules continuously react in both directions but with no overall change in composition.

• Equilibrium occurs when forward rate = reverse rate.
• Concentrations of reactants and products remain constant at equilibrium.
• Equilibrium is a dynamic state with ongoing reactions in both directions.

For equilibrium to be established, the system must be closed to prevent loss or gain of substances.

Temperature, pressure, and concentration affect the rates of forward and reverse reactions but do not prevent equilibrium from being reached if the system remains closed.

If the system is open, reactants or products can escape, preventing the concentrations from stabilising and equilibrium from being established.

• System must be closed to establish equilibrium.
• External changes affect rates but equilibrium can still be reached.
• Open systems do not maintain constant concentrations; no equilibrium.

Consider the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) in a closed container.

Initially, only nitrogen and hydrogen gases are present, so the forward reaction rate is high and the reverse rate is zero.

As ammonia forms, its concentration increases, increasing the reverse reaction rate until it equals the forward rate, establishing equilibrium where concentrations of N₂, H₂, and NH₃ remain constant over time.

• Forward and reverse reaction rates become equal at equilibrium.
• Concentrations of all species remain constant but not necessarily equal.
• The reaction continues dynamically in both directions at equilibrium.

In a reversible reaction, the system can reach a state of dynamic equilibrium where the forward and reverse reaction rates are equal, and the concentrations of reactants and products remain constant. The position of equilibrium refers to the relative concentrations of reactants and products at equilibrium. This position can be influenced by changes in temperature, pressure, concentration, and the presence of catalysts. Understanding these factors allows prediction of how a system responds to changes, based on Le Chatelier's principle, which states that a system at equilibrium will adjust to counteract any imposed change.

• Equilibrium is dynamic: forward and reverse rates equal.
• Position of equilibrium indicates relative amounts of reactants and products.
• Le Chatelier's principle predicts shifts in equilibrium position after changes.

Temperature affects both the position of equilibrium and the rates of the forward and reverse reactions. Every reversible reaction involves either absorption or release of heat (endothermic or exothermic). Increasing temperature favours the endothermic direction, as the system absorbs the added heat, shifting equilibrium to produce more products or reactants accordingly. Conversely, decreasing temperature favours the exothermic direction.

For example, consider the reversible reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol (exothermic forward reaction).

Increasing temperature shifts equilibrium to the left (towards reactants), reducing ammonia yield. Decreasing temperature shifts equilibrium to the right (towards ammonia), increasing yield but slowing reaction rate. Therefore, industrial processes balance temperature to optimise yield and rate.

• Increasing temperature shifts equilibrium towards endothermic side.
• Decreasing temperature shifts equilibrium towards exothermic side.
• Temperature changes affect reaction rates: higher temperature increases rates of both directions.
• Example: Haber process equilibrium shifts left with temperature increase due to exothermic ammonia formation.

Pressure changes affect equilibrium only when gases are involved and the number of moles of gas differs between reactants and products. Increasing pressure favours the side with fewer moles of gas, reducing total pressure. Decreasing pressure favours the side with more moles of gas.

For example, in the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

There are 4 moles of gas on the left and 2 moles on the right. Increasing pressure shifts equilibrium to the right, increasing ammonia production. Decreasing pressure shifts equilibrium to the left, reducing ammonia yield.

• Pressure changes affect equilibrium only for gaseous reactions with unequal moles.
• Increasing pressure shifts equilibrium to side with fewer gas molecules.
• Decreasing pressure shifts equilibrium to side with more gas molecules.
• Example: Haber process equilibrium shifts right with increased pressure due to fewer moles of NH₃ gas.

Changing the concentration of reactants or products disturbs the equilibrium. According to Le Chatelier's principle, the system shifts to oppose the change by consuming the added substance or replacing the removed substance.

For example, in the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Adding more SO₂ shifts equilibrium to the right, producing more SO₃. Removing SO₃ shifts equilibrium to the right to replace it. Similarly, decreasing O₂ concentration shifts equilibrium to the left to produce more O₂.

• Increasing concentration of a reactant or product shifts equilibrium to consume it.
• Decreasing concentration shifts equilibrium to produce more of the removed substance.
• Equilibrium shifts to restore balance in concentrations.
• Example: Contact process equilibrium shifts right when SO₂ concentration increases.

Catalysts increase the rate of both forward and reverse reactions equally by lowering the activation energy. They do not affect the position of equilibrium or the equilibrium constant. Instead, catalysts help the system reach equilibrium faster.

For example, in the Haber process, an iron catalyst speeds up ammonia formation and decomposition but does not change the equilibrium concentrations of reactants and products.

• Catalysts speed up forward and reverse reactions equally.
• They do not change equilibrium position or yield.
• Catalysts reduce activation energy, increasing reaction rates.
• Example: Iron catalyst in Haber process accelerates reaching equilibrium without shifting it.

The Haber process is an industrial method used to synthesise ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. Ammonia is a vital chemical used primarily to manufacture fertilisers, which support global food production. The process combines nitrogen from the air and hydrogen typically derived from natural gas under specific conditions to produce ammonia efficiently.

• Ammonia is essential for fertiliser production.
• Nitrogen is obtained from air (approximately 78% nitrogen).
• Hydrogen is usually sourced from methane (natural gas).

The chemical reaction in the Haber process can be represented by a balanced symbol equation showing the reactants and products with their physical states:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) This equation indicates that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to form two molecules of ammonia gas.

The double arrow (⇌) signifies that the reaction is reversible, meaning ammonia can decompose back into nitrogen and hydrogen under certain conditions.

• Reactants: nitrogen gas (N₂) and hydrogen gas (H₂).
• Product: ammonia gas (NH₃).
• Balanced equation: N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
• Reaction is reversible, indicated by ⇌.

The Haber process reaction is exothermic, releasing heat as ammonia forms. Because the reaction is reversible, the position of equilibrium depends on temperature, pressure, and catalysts used. Industrially, the process is conducted at approximately 450 °C and a pressure of about 200 atmospheres (20,000 kPa) with an iron catalyst to optimise ammonia yield and rate.

The high pressure favours the formation of ammonia because there are fewer gas molecules on the product side (2 molecules) than on the reactant side (4 molecules), according to Le Chatelier's principle.

The moderate temperature balances the rate of reaction and the equilibrium yield since lower temperatures favour ammonia formation but slow the reaction rate.

• Reaction is exothermic.
• Typical conditions: 450 °C, 200 atm pressure, iron catalyst.
• High pressure favours ammonia formation due to fewer gas molecules on product side.
• Temperature chosen to balance rate and yield.

Given the reactants nitrogen and hydrogen gases, write the balanced symbol equation for ammonia production.

Step 1: Write the unbalanced equation: N₂(g) + H₂(g) → NH₃(g)

Step 2: Balance nitrogen atoms: 1 N₂ molecule contains 2 nitrogen atoms, so place 2 NH₃ molecules to have 2 nitrogen atoms on the product side: N₂(g) + H₂(g) → 2 NH₃(g)

• Step 3: Balance hydrogen atoms: 2 NH₃ molecules contain 6 hydrogen atoms, so place 3 H₂ molecules on the reactant side: N₂(g) + 3 H₂(g) → 2 NH₃(g)
• Step 4: Confirm balanced atoms: Nitrogen: 2 on both sides; Hydrogen: 6 on both sides.
• Final balanced equation: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

The Haber process is an industrial method for synthesising ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. Ammonia is a key raw material for fertilisers and other chemicals. The process requires a continuous supply of pure nitrogen and hydrogen gases, which are obtained from different sources.

The overall balanced equation for the Haber process is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (with conditions: 450 °C, 200 atm, iron catalyst)

• Ammonia synthesis requires nitrogen and hydrogen gases.
• Nitrogen and hydrogen must be sourced industrially in large quantities.

Hydrogen used in the Haber process is primarily produced from methane (CH₄), the main component of natural gas. The industrial method is called steam reforming, which involves reacting methane with steam (water vapour) at high temperature (about 700–1100 °C) in the presence of a nickel catalyst.

The main reaction is: CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)

This reaction is endothermic and produces carbon monoxide and hydrogen gas. To maximise hydrogen yield, the carbon monoxide is further reacted with steam in the water-gas shift reaction: CO(g) + H₂O(g) → CO₂(g) + H₂(g)

• Steam reforming converts methane and steam into hydrogen and carbon monoxide.
• Water-gas shift reaction converts carbon monoxide and steam into carbon dioxide and more hydrogen.
• Hydrogen gas is then purified by removing carbon dioxide and other impurities.

Nitrogen is obtained from the air, which contains approximately 78% nitrogen and 21% oxygen, with small amounts of other gases. The air is first filtered and compressed, then cooled to very low temperatures to liquefy it.

Fractional distillation of liquid air separates nitrogen from oxygen and other gases based on their different boiling points. Nitrogen boils at −196 °C, oxygen at −183 °C. By careful warming, nitrogen gas is collected as it evaporates first.

The purified nitrogen gas is then fed into the Haber process reactor.

• Air is liquefied by cooling and compressed to separate gases.
• Fractional distillation separates nitrogen from oxygen due to different boiling points.
• Nitrogen gas is collected and purified for use in ammonia synthesis.

The key chemical reactions for hydrogen production are:

1. Steam reforming: CH₄(g) + H₂O(g) → CO(g) + 3H₂(g) (high temperature, nickel catalyst)

2. Water-gas shift: CO(g) + H₂O(g) → CO₂(g) + H₂(g) (moderate temperature)

• Hydrogen is produced from methane and steam in two steps.
• Nitrogen is separated from air by fractional distillation.
• Both gases must be purified before entering the Haber process.

The Haber process is an industrial method for producing ammonia (NH₃) by reacting nitrogen (N₂) from the air with hydrogen (H₂), usually derived from methane. The balanced chemical equation is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ/mol (exothermic)

This reaction is reversible and reaches a dynamic equilibrium where the forward and reverse reaction rates are equal. The process requires specific conditions to maximise ammonia yield and maintain a practical reaction rate.

• Balanced equation: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• Reaction is exothermic and reversible

The reaction is exothermic, so according to Le Chatelier's principle, lowering temperature favours ammonia formation. However, at low temperatures, the reaction rate is too slow for industrial use. Therefore, a compromise temperature of about 450 °C is used to balance a reasonable rate with acceptable yield.

• Lower temperature favours ammonia formation (exothermic reaction)
• Too low temperature results in slow reaction rate
• 450 °C is a compromise between rate and yield

The reaction involves 4 moles of gas (1 N₂ + 3 H₂) producing 2 moles of NH₃. Increasing pressure shifts equilibrium towards fewer gas molecules, favouring ammonia production. Industrially, a high pressure of about 20,000 kPa (200 atm) is applied to increase yield. However, very high pressures require stronger, more expensive equipment and increase safety risks.

• High pressure favours ammonia formation (fewer gas molecules)
• Typical pressure: 20,000 kPa (200 atm)
• High pressure increases yield but raises costs and safety concerns

An iron catalyst is used to increase the rate at which equilibrium is reached without affecting the position of equilibrium or the yield. The catalyst provides an alternative reaction pathway with lower activation energy, speeding up both forward and reverse reactions equally.

• Iron catalyst speeds up reaction rate
• Does not change equilibrium position or yield
• Allows equilibrium to be reached faster

The Haber process conditions are carefully chosen to optimise ammonia production:

Temperature of 450 °C balances reaction rate and equilibrium yield.

Pressure of 20,000 kPa (200 atm) shifts equilibrium towards ammonia, increasing yield but requiring robust equipment and safety measures.  - Iron catalyst accelerates reaction rate, reducing time to reach equilibrium without affecting yield.

• 450 °C temperature
• 20,000 kPa (200 atm) pressure
• Iron catalyst

The Contact process is an industrial method used to produce sulfuric acid, one of the most important chemicals worldwide. A crucial step in this process is the conversion of sulfur dioxide (SO₂) to sulfur trioxide (SO₃). Sulfur trioxide is then used to produce sulfuric acid by reaction with water.

Sulfur dioxide is obtained by burning sulfur or roasting sulfide ores, and oxygen is supplied from air. The conversion of SO₂ to SO₃ is a reversible reaction and is carefully controlled to maximise yield and efficiency.

• Sulfur dioxide (SO₂) is oxidised to sulfur trioxide (SO₃).
• This step is essential for sulfuric acid manufacture.
• The reaction is reversible and reaches equilibrium under certain conditions.

The chemical reaction converting sulfur dioxide to sulfur trioxide is represented by the balanced symbol equation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g). This shows that two molecules of sulfur dioxide react with one molecule of oxygen to form two molecules of sulfur trioxide.

The symbol ⇌ indicates that the reaction is reversible, meaning the forward and backward reactions occur simultaneously and can reach a state of dynamic equilibrium.

• Reactants: sulfur dioxide (SO₂) and oxygen (O₂), both gases.
• Products: sulfur trioxide (SO₃), a gas under reaction conditions.
• The reaction is balanced with 2 SO₂ molecules reacting with 1 O₂ molecule to form 2 SO₃ molecules.

The conversion of SO₂ to SO₃ is carried out under specific conditions to optimise the rate and yield of sulfur trioxide. The typical conditions are:

Temperature: approximately 450 °C. This temperature balances the rate of reaction and the position of equilibrium, as the reaction is exothermic.

Pressure: around 200 kPa (2 atm). Increased pressure favours the formation of fewer gas molecules, shifting equilibrium towards SO₃ production, but very high pressures are not economically viable or safe in industry.

• High temperature (about 450 °C) to increase reaction rate.
• Moderate pressure (about 200 kPa) to favour product formation.
• Use of a catalyst to speed up the reaction without changing equilibrium position.

Vanadium(V) oxide (V₂O₅) is used as a catalyst in the Contact process. It increases the rate at which equilibrium is reached by providing an alternative reaction pathway with lower activation energy.

The catalyst does not affect the position of equilibrium or the equilibrium constant; it only helps the reaction to reach equilibrium faster.

This is important industrially to maximise production efficiency and reduce costs.

• Vanadium(V) oxide (V₂O₅) is the catalyst.
• Catalyst lowers activation energy, increasing reaction rate.
• Catalyst does not change equilibrium yield, only speed of reaching equilibrium.

The reaction is reversible, meaning sulfur trioxide can decompose back into sulfur dioxide and oxygen. At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant.

Industrial conditions are chosen to maximise the yield of sulfur trioxide while maintaining a reasonable reaction rate and safety.

Sulfur trioxide produced is then used to make sulfuric acid by reaction with water, an important industrial chemical used in fertilisers, detergents, and many other products.

• Reversible reaction reaches dynamic equilibrium.
• Conditions balance yield, rate, safety, and cost.
• Sulfur trioxide is an intermediate for sulfuric acid production.

The Contact process is an industrial method used to manufacture sulfuric acid (H₂SO₄). A key step in this process is the production of sulfur dioxide (SO₂), which then reacts with oxygen (O₂) to form sulfur trioxide (SO₃). Understanding the sources of SO₂ and O₂ is essential for grasping the overall process.

• Sulfur dioxide is a reactant formed before sulfur trioxide in the Contact process.
• Oxygen is required to oxidise sulfur dioxide to sulfur trioxide.
• Both gases must be obtained efficiently and economically for industrial use.

Elemental sulfur can be burned in air to produce sulfur dioxide. This is a direct combustion reaction where sulfur reacts with oxygen from the air.

• Reaction: S(s) + O₂(g) → SO₂(g)
• Balanced equation: S₈(s) + 8O₂(g) → 8SO₂(g) (using elemental sulfur as S₈)
• Conditions: Burning sulfur in excess air at atmospheric pressure.
• Observation: A blue flame is often seen, and sulfur dioxide gas is colourless with a choking smell.
• This method provides a pure source of SO₂ gas for the Contact process.

Many metal sulfide ores contain sulfur combined with metals, such as iron pyrite (FeS₂). Roasting these ores in air produces sulfur dioxide and metal oxides.

• Example reaction: 4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)
• This is a thermal oxidation reaction where sulfide ions are oxidised to sulfur dioxide gas.
• Conditions: Heating the ore strongly in the presence of air.
• Observation: The ore glows red-hot, and sulfur dioxide gas is released.
• This method utilises naturally occurring sulfide minerals as a sulfur source.

Oxygen required for the Contact process is obtained from atmospheric air, which contains approximately 21% oxygen by volume.

• Air is filtered and compressed to supply oxygen for the reaction.
• No chemical separation is needed for the Contact process; air is used directly.
• Oxygen reacts with sulfur dioxide over a catalyst to form sulfur trioxide.
• The presence of nitrogen and other gases in air does not interfere significantly with the process.

The key chemical reactions for producing sulfur dioxide and oxygen sources in the Contact process are summarised below.

• Burning sulfur: S(s) + O₂(g) → SO₂(g)
• Roasting iron pyrite: 4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)
• Oxygen source: O₂(g) from air (approximately 21% by volume)
• Conditions: Combustion or roasting at elevated temperatures in air.
• Sulfur dioxide and oxygen gases are then used in the catalytic oxidation step of the Contact process.

The Contact process is an industrial method used to produce sulfur trioxide (SO₃) by oxidising sulfur dioxide (SO₂). Sulfur trioxide is a key intermediate in the manufacture of sulfuric acid, one of the most important industrial chemicals.

The overall reaction is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

• Sulfur dioxide is obtained by burning sulfur or roasting sulfide ores.
• Oxygen is supplied from air.
• The reaction is reversible and exothermic.

The balanced symbol equation for the Contact process is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

This reaction is reversible and reaches equilibrium under certain conditions.

Because the forward reaction is exothermic, lowering temperature favours SO₃ formation, but too low temperature slows the reaction rate significantly.

• Forward reaction: 2SO₂ + O₂ → 2SO₃ (exothermic)
• Reversible reaction indicated by ⇌
• Equilibrium position depends on temperature and pressure

The industrial conditions chosen balance reaction rate and equilibrium yield to maximise sulfur trioxide production efficiently and economically.

The typical conditions are: temperature of 450 °C, pressure of 200 kPa (2 atm), and the use of vanadium(V) oxide (V₂O₅) as a catalyst.

Temperature: 450 °C is a compromise between a high rate of reaction and a favourable equilibrium position. Lower temperatures increase SO₃ yield but slow the reaction; higher temperatures increase rate but reduce yield due to equilibrium shift.

• Temperature: 450 °C
• Pressure: 200 kPa (2 atm)
• Catalyst: Vanadium(V) oxide (V₂O₅)

Increasing pressure favours the forward reaction because the number of gas molecules decreases from 3 (2SO₂ + O₂) to 2 (2SO₃).

A moderate pressure of 200 kPa (2 atm) is used to increase yield without incurring excessive costs or safety risks associated with very high pressures.

Higher pressure increases the rate and shifts equilibrium towards SO₃, but very high pressures are not economically viable.

• Forward reaction reduces total gas moles from 3 to 2
• Moderate pressure (2 atm) balances yield and cost
• Excessive pressure increases plant costs and safety hazards

Vanadium(V) oxide (V₂O₅) acts as a heterogeneous catalyst, providing an alternative reaction pathway with lower activation energy.

The catalyst increases the rate at which equilibrium is reached but does not change the position of equilibrium or the equilibrium constant.

It allows the reaction to proceed efficiently at 450 °C, which is a practical temperature for industrial operation.

• V₂O₅ provides surface for adsorption of reactants
• Lowers activation energy, increasing reaction rate
• Does not affect equilibrium position

The Haber process synthesises ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. The balanced chemical equation is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ/mol (exothermic)

The Contact process converts sulfur dioxide (SO₂) and oxygen (O₂) into sulfur trioxide (SO₃), a key step in sulfuric acid production. The balanced equation is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = -198 kJ/mol (exothermic)

• Both reactions are reversible and exothermic.
• Equilibrium position and reaction rate depend on temperature, pressure, and catalysts.
• Industrial conditions balance yield, rate, safety, and cost.

Increasing temperature generally increases reaction rates by providing more kinetic energy to reactant molecules, increasing collision frequency and energy.

However, for exothermic reactions, increasing temperature shifts the equilibrium position towards reactants, reducing product yield (Le Chatelier's principle).

In the Haber process, a moderate temperature (~450 °C) is chosen to balance a reasonable reaction rate with acceptable ammonia yield. Lower temperatures increase yield but slow the reaction excessively, making the process uneconomical. Higher temperatures increase rate but reduce yield. Similarly, in the Contact process, a temperature around 450 °C is used to optimise the rate and equilibrium position for SO₃ production.

• Temperature increase → faster reaction rate but equilibrium shifts to reactants for exothermic reactions.
• Moderate temperature balances rate and yield in both processes.
• Excessively high temperatures reduce product yield despite faster rates.

Increasing pressure favours the side of the reaction with fewer gas molecules, shifting equilibrium towards products in both processes.

In the Haber process, 1 mole of N₂ reacts with 3 moles of H₂ (4 moles total) to form 2 moles of NH₃, so high pressure shifts equilibrium towards ammonia.

High pressure also increases collision frequency, increasing reaction rate. Industrially, pressures around 20,000 kPa (200 atm) are used in the Haber process to maximise ammonia yield and rate economically and safely, considering equipment costs and safety risks of very high pressures. In the Contact process, pressure is kept relatively low (~200 kPa or 2 atm) because increasing pressure has less effect on equilibrium (4 moles reactants to 2 moles products) and higher pressures increase costs and safety risks without significant yield improvement.

• High pressure favours product formation when fewer gas molecules are produced.
• Haber process uses very high pressure (~200 atm) to increase ammonia yield and rate.
• Contact process uses moderate pressure (~2 atm) balancing yield, rate, and safety.

Catalysts provide an alternative reaction pathway with lower activation energy, increasing reaction rate without affecting equilibrium position or yield.

In the Haber process, an iron catalyst is used to increase the rate of ammonia formation at moderate temperature and high pressure.

In the Contact process, vanadium(V) oxide (V₂O₅) acts as a catalyst to speed up the conversion of SO₂ to SO₃ at around 450 °C and moderate pressure, improving efficiency and reducing energy consumption.

• Catalysts increase reaction rate but do not change equilibrium position.
• Iron catalyst in Haber process speeds ammonia synthesis.
• Vanadium(V) oxide catalyst in Contact process speeds SO₃ formation.

Extremely high pressures and temperatures increase equipment costs, energy consumption, and safety risks such as explosions or leaks.

The Haber process uses high pressure (~200 atm) because the increase in ammonia yield and rate justifies the cost and safety measures.

Temperatures are kept moderate (~450 °C) to balance energy costs and yield, avoiding excessive energy consumption and catalyst damage at higher temperatures. The Contact process operates at lower pressure (~2 atm) to reduce equipment costs and safety risks, as pressure has less effect on equilibrium here. Temperature is optimised (~450 °C) to balance rate and yield while preserving catalyst life and minimising side reactions. Overall, industrial conditions represent a compromise between maximising product yield and rate, minimising energy and equipment costs, and ensuring safe operation.

• High pressure and temperature increase costs and safety risks.
• Conditions chosen to optimise yield, rate, and economic viability.
• Catalyst use reduces energy consumption by increasing rate at moderate conditions.